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This question already has an answer here:

Is there a command on Mathematica that helps me to get the answer of some harmonic series in terms on $\zeta(2n)$ instead of $\pi^{2n}$? Let me give you an example:

The command :

Sum[HarmonicNumber[n, 5]/n^8, {n, 1, Infinity}]

gives :

$$-(1/63) [\pi]^6 \zeta[7] - (13/15) [\pi]^4 \zeta[9] - 55 [\pi]^2 \zeta[11] + 644 \zeta[13]$$

I know converting $\pi^2$, $\pi^4$ and $\pi^6$ to $\zeta(2)$ , $\zeta(4)$ and $\zeta(6)$ is not a big deal but I deal with harmonic series a lot and converting is time consuming. Also I deal with harmonic series of high height which means more converting to do. So is there any command that gives the answer of the harmonic sum in terms of $\zeta(2n)$ instead of $\pi^{2n}$?

Thank you.

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marked as duplicate by J. M. will be back soon Nov 13 at 9:54

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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One can make use of Simplify with Assumptions

I. Compute the sum

s=Sum[HarmonicNumber[n,5]/n^8,{n,1,Infinity}]
(* -(1/63) π^6 Zeta[7]-13/15 π^4 Zeta[9]-55 π^2 Zeta[11]+644 Zeta[13] *)

II. Make a table of Zeta-functions with even arguments

t=Flatten[Table[{ζ[2n]==Zeta[2n]},{n,0,6}]]
(* {ζ[0]==-(1/2),ζ[2]==π^2/6,ζ[4]==π^4/90,ζ[6]==π^6/945,ζ[8]==π^8/9450,ζ[10]==π^10/93555,ζ[12]==(691 π^12)/638512875} *)

III. Simplify with assumptions

Simplify[s/.{π->x,Zeta[n_]->ζ[n]},Assumptions->t/.{π->x}]
(* -15 ζ[6] ζ[7]-78 ζ[4] ζ[9]-330 ζ[2] ζ[11]+644 ζ[13] *)

Comments:

  1. A much cleaner solution would be to use the method of Carl Woll to deactivate the Zetafunction. However, it only works if the sum can be computed in the desired terms, i.e., in terms of Zeta[2n+1] and Zeta[2n]. This seems not to be the case.
  2. The present method only replaces the occurrences of $\pi^{2n}$ for $n\ge1$. The zeroth-order term is left unchanged. Below I show how Zeta[0] can be taken into account.

Different approach using SolveAlways

Define some functions

Clear[sf,sx,sxx]
sf[a_,b_]:=Sum[HarmonicNumber[n,a]/n^b,{n,1,Infinity}]/.{Zeta[n_]->ζ[n]}
sxx[c_]:=Sum[a[i]ζ[i]Zeta[c-i],{i,c,Floor[c/2],-2}]
sx[c_]:=Sum[a[i]ζ[i]ζ[c-i],{i,c,Floor[c/2],-2}]

Let us try

k=5;l=8;
sx[k+l]/.First[SolveAlways[(sf[k,l]==sxx[k+l]),Table[ζ[i],{i,k+l,Floor[(k+l)/2],-2}]]]
(* -15 ζ[6] ζ[7]-78 ζ[4] ζ[9]-330 ζ[2] ζ[11]-1288 ζ[0] ζ[13] *)
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